$\int \sin (x) \cos (x) dx$ , $\int 1 / (x \ln x) dx$ , $\int x e^{x^{2}} dx$

By Anonymous on Sunday, November 26, 2000 - 08:04 pm :

Hi,

I'm not sure what the following integrates to and not sure whether my method is correct.

\int \sin x \cos x dx

So we know that Sin2x = 2SinxCosx, hence SinxCosx = 0.5(Sin2x), is it ok so far?
hence, I think,

\frac{1}{2} \int \sin 2 x dx = \frac{1}{2} \times (\frac{1}{2} \times \cos 2 x)

???
Is this right? I am not too sure here.

Thanks for any help in advance

By James Myatt (Jem50) on Sunday, November 26, 2000 - 08:18 pm :

Not quite. You have got the right idea and it's correct except that you've integrated sinx to cosx, where as it's actually -cosx, and you've forgotten the constant of integration. Nevertheless, that is the correct method.

An alternative method is to substitute u = sinx, and you'll find that it's quite a neat substitution. But it still gives the same answer as the way you've done it (nearly) and that way is easier.

James

By Anonymous on Sunday, November 26, 2000 - 08:51 pm :

Thanks James, so the answer of

\int \sin x \cos x dx = - \frac{1}{2} \times (\frac{1}{2} \times \cos 2 x) + c

, is this correct?
Thanks

By Anonymous on Sunday, November 26, 2000 - 08:52 pm :

Yes, the final answer is -1/4 cos(2x) + C.

By Anonymous on Sunday, November 26, 2000 - 09:11 pm :

How do you go about integrating the following?

\int 1 / (x \times \ln x) dx

Do I use a sub u=xlnx??

By Barkley Bellinger (Bb246) on Sunday, November 26, 2000 - 09:17 pm :

The integrand can be written as
(1/x) / (lnx) , so the answer is ln(lnx) + const.

By Michael Doré (Md285) on Sunday, November 26, 2000 - 09:17 pm :

You may prefer to do that with a substitution rather than by inspection:

Try $u = \ln x$

$du / dx = 1 / x$

So $dx / du = x$

Therefore our integral is:

$\int [1 / (x u) dx / du du] = \int [x / (x u) du] = \int [1 / u du] = \ln u + C = \ln (\ln x) + C$

By Anonymous on Sunday, November 26, 2000 - 10:41 pm :

Thanks Barkley & Michael!

Why does

\int x e^{x^{2}} dx = e^{x^{2}} / 2 + c

?, can anyone explain it to me through using substitution or other easy to understand methods. Thanks.

By James Lingard (Jchl2) on Sunday, November 26, 2000 - 10:47 pm :

With the substitution

u = x^{2}

, you get

du / dx = 2 x

, so

dx / du = 1 / (2 x)

, so the integral becomes

$\int x \times e^{u} / (2 x) du = (1 / 2) \int e^{u} du = (1 / 2) e^{u} + c$

James.

By Anonymous on Sunday, November 26, 2000 - 11:16 pm :

Great! thanks James.

By Barkley Bellinger (Bb246) on Monday, November 27, 2000 - 10:50 am :

Alternatively you can spot that the differential of the suggested answer (using the chain rule) is what you are integrating.